Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 6 - Section 6.6 - Variation - 6.6 Exercises - Page 418: 35

Answer

$x=27$

Work Step by Step

Recall: If $y$ varies directly as $x$ then, $y=kx$ where $k$ is the constant of variation. Since $x=9$ w3&=hen $y=3$, then: \begin{align*} y&=kx\\\\ 3&=k(9)\\\\ \frac{3}{9}&=\frac{k(9)}{9}\\\\ \frac{1}{3}&=k \end{align*} Thus, the direct variation's equation is: $$y=\frac{1}{3}x$$ To find the value of $x$ when $y=9$, substitute these values into the equation above to obtain: \begin{align*} y&=\frac{1}{3}x\\\\ 9&=\frac{1}{3}x\\\\ 3(9)&=\frac{1}{3}x \cdot 3\\\\ 27&=x \end{align*}
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